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Thread: trig function derivative

  1. #1
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    trig function derivative

    Given that $\displaystyle x = cosy^{2} $

    find dy/dx in terms of y;

    I am not sure which rule to use, do I use the product rule or chain rule?

    I thought I should use the chain rule and this is what I get;

    first differentiate cos which = -sin

    than do y^2 to get 2y.

    so $\displaystyle \frac{dx}{dy} = -sin2y $

    I know this wrong, but can someone please tell me what I am doing wrong?
    the correct answer is $\displaystyle \frac{dx}{dy} = -siny^{2} \times 2y $


    thanks,
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  2. #2
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    Quote Originally Posted by Tweety View Post
    Given that $\displaystyle x = cosy^{2} $

    find dy/dx in terms of y;

    I am not sure which rule to use, do I use the product rule or chain rule?

    I thought I should use the chain rule and this is what I get;

    first differentiate cos which = -sin

    than do y^2 to get 2y.

    so $\displaystyle \frac{dx}{dy} = -sin2y $

    I know this wrong, but can someone please tell me what I am doing wrong?
    the correct answer is $\displaystyle \frac{dx}{dy} = -siny^{2} \times 2y $


    thanks,
    which do you mean ...

    $\displaystyle x = \cos(y^2)$ or $\displaystyle x = (\cos{y})^2$

    ?
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  3. #3
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    Quote Originally Posted by skeeter View Post
    which do you mean ...

    $\displaystyle x = \cos(y^2)$ or $\displaystyle x = (\cos{y})^2$

    ?
    The first one. $\displaystyle x = \cos(y^2)$
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  4. #4
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    Quote Originally Posted by Tweety View Post
    The first one. $\displaystyle x = \cos(y^2)$
    your question says to find $\displaystyle \frac{dy}{dx}$, but the answer is $\displaystyle \frac{dx}{dy}$ , so I'll do both.

    $\displaystyle \frac{d}{dx}[x = \cos(y^2)]$

    chain rule on the right side ...

    $\displaystyle 1 = -\sin(y^2) \cdot 2y \frac{dy}{dx}$

    $\displaystyle \frac{dy}{dx} = -\frac{1}{2y\sin(y^2)}$



    $\displaystyle \frac{d}{dy}[x = \cos(y^2)]$

    $\displaystyle \frac{dx}{dy} = -\sin(y^2) \cdot 2y$

    $\displaystyle \frac{dx}{dy} = -2y \sin(y^2)$
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  5. #5
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    I think am not understanding how your using the chain rule on $\displaystyle cosy^{2} $

    I use this expression ;
    $\displaystyle y = f(x)^{n} = \frac{dy}{dx} = nf(x)^{n-1} f '(x) $

    there's no 'inside bit in the expression $\displaystyle cosy^{2}$ its just one, so how do you separate the variables?

    for example in this expression $\displaystyle y = 6(4x-2)^{3} $ I would first differenciate the expression in the bracket and than do the out side leaving the inside the same and than multiply both expressions.

    So am just wondering how $\displaystyle cosy^{2} $ fits into the above formula.

    Thank you.
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  6. #6
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    note that for $\displaystyle x = \cos(y^2)$ , only the y is squared, not the cosine function.

    the derivative w/r to x is

    $\displaystyle 1 = -\sin(y^2) \cdot (the \, derivative \, of \, y^2) = -sin(y^2) \cdot 2y \cdot \frac{dy}{dx}$
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